Unit 4: Barking Deer Foraging Preferences

UNIT 4 LEARNING JOURNAL Part 1 Barking Deer Foraging Preferences (a) Hypotheses: ●​ Null hypothesis (H₀): There is no difference in the average scores of students in the reading and writing exams. H0:μread−μwrite=0 ●​ Alternative hypothesis (H₁): There is a difference in the average scores of students in the reading and writing exams. H1:μread−μwrite≠0 (b) Prerequisites for carrying out the paired t-test: We look for the following circumstances: Randomness: This requirement is met because the 250-student sample was chosen at random. Independence: Because every student submits a pair of reading and writing scores, score differences are unaffected by those of other students. Normality of differences: Given the large sample size (n = 250), the normality criteria looks fair, since the histogram of differences appears to be approximately symmetric and unimodal. Everything appears to be in order! (c) T-test calculations: The formula for the t-test statistic for paired data is: where: ●​ xread-write = − 0.545 (mean difference) ●​ s = 8.887 (standard deviation of differences) ●​ n = (sample size) so t =​≈ − 0.97 Degrees of freedom (df):​ The degrees of freedom for a paired t-test is: df = n − 1 = 250 − 1 = 249 Conclusion:​ The p-value = 0.39 means there is a 39% probability of observing a difference this extreme or more if the null hypothesis is true. Since the p-value is much larger than the typical significance level of 0.05, we fail to reject the null hypothesis. There is no convincing evidence that there is a difference between the average reading and writing scores. (d) Potential error The error we might have made is a Type II error – failing to reject the null hypothesis when there actually is a difference in reading and writing scores. In this situation, it would entail concluding that there is no difference between the average scores, even while a real difference does exist. (e) Confidence interval The average score difference's confidence interval is probably going to include 0 because the p-value is high and the null hypothesis was not successfully rejected. This supports the notion that there is no discernible difference between the two scores by implying that the true mean difference might be zero. Part 2 (1) State the hypotheses: ●​ Null hypothesis (H₀): There is no difference in average city mileage between automatic and manual transmission cars. H0 : μautomatic = μmanual ●​ Alternative hypothesis (H₁): There is a difference in average city mileage between automatic and manual transmission cars. H1 : μautomatic ≠ μmanual (2) Calculate the t-statistic: The formula for the t-statistic for two independent samples is: Where: ●​ xˉ1 = 16.12 (mean city MPG for automatic cars) ●​ s1 = 3.58 (SD for automatic cars) ●​ n1 = 26 (sample size for automatic cars) ●​ xˉ2 = 19.85 (mean city MPG for manual cars) ●​ s2 = 4.5 (SD for manual cars) ●​ n2 = 26 (sample size for manual cars) (3) Degrees of freedom: Using the Welch–Satterthwaite equation: (4) Conclusion: Given the p-value = 0.0029 and using a significance level of α = 0.05: ●​ The p-value is much smaller than 0.05. ●​ Therefore, we reject the null hypothesis. Conclusion: There is strong evidence that the average city mileage of manual and automatic cars is significantly different. REFERENCES: Moore, D. S., McCabe, G. P., Craig, B. A., & Alwan, L. C. (2021). The practice of statistics for business and economics (5th ed.). W. H. Freeman.